State De Moivre's theorem and prove it for all integer values?

3 Answers
May 3, 2018

#(cos theta + i sin theta)^n = cosn theta + isinntheta #

Proof by induction

Explanation:

#(cos theta + i sin theta)^n = cosn theta + isinntheta #

Prove by induction:

Basis case: #n = 1 #

#=> ( cos theta + isin theta)^1 = cos theta + i sin theta #

Hence basis case holds

Asume true for #n=k #

#(cos theta + i sin theta)^k = cosk theta + isinktheta #

Showing holds for # n = k+1#

#(cos theta + i sin theta)^(k+1) = (cosk theta + isinktheta)(cos theta + isin theta) #

#(cos theta + i sin theta)^(k+1) = #

#= costheta cosk theta + isinthetacosktheta + isinkthetacostheta - sinktheta sintheta #

# = costhetacosktheta - sinkthetasintheta + i ( sinthetacosktheta + sinktheta cos theta ) #

#=> = cos ( (k+1) theta ) + i sin ( (k+1) theta) #

Hence it holds true for #k+1#

It holds for basis case #n=1# and #n = k + 1 # for all #k # that holds, hence holds for #n =1 #, #n =2 # , #n=3# etc...

#=> "holds" AA n in ZZ #

May 3, 2018

# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR#

Explanation:

De Moivre's Theorem, states that:

# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR#

We seek to validate this result for #n in NN#, which we can prove using Mathematical Induction:

Induction Proof - Hypothesis

We seek to prove that:

#

# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \ # ..... [A]

So let us test this assertion, [A], using Mathematical Induction:

Induction Proof - Base case:

Consider, the special case #n=1# in which case we have the trivial result:

# (cos theta+i sin theta)^1 = cos(1 theta) + isin(1 theta) #

So the given result is true when #n=1#.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when #n=m#, for some #m in NN, m gt 1#, in which case for this particular value of #m# we have:

# (cos theta+i sin theta)^m = cos(m theta) + isin(m theta) \ \ \ # ..... [B]

Then, consider the expression:

# E = (cos theta+i sin theta)^(m+1) #

# \ \ \ = (cos theta+i sin theta)^m(cos theta+i sin theta) #

# \ \ \ = (cos(m theta) + isin(m theta))(cos theta+i sin theta) \ \ # (using [B])

# \ \ \ = cos(m theta)cos theta + isin(m theta)cos theta + cos(m theta)i sin theta + isin(m theta)i sin theta #

# \ \ \ = {cos(m theta)cos theta -sin(m theta) sin theta} + i{sin(m theta)cos theta + icos(m theta) sin theta } #

# \ \ \ = cos(m+1) theta + isin(m+1) theta #

Which is the given result [A] with #n=m+1#

Induction Proof - Summary

So, we have shown that if the given result [A] is true for #n=m#, then it is also true for #n=m+1#. But we initially showed that the given result was true for #n=1# so it must also be true for #n=2, n=3, n=4, ... # and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for #n in NN#

Hence we have:

# (cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \ # QED

May 3, 2018

See below.

Explanation:

First let us prove #r[costheta+jsintheta]#=#re^[jtheta]#.

Let #z=costheta+jsintheta...........#[1]##, therefore, #dz/[d theta#=#-[sintheta-jcostheta]#

=#j^2sintheta+jcostheta# = #j[costheta+jsintheta]#[ since#j^2=-1]#

But #j[cos theta+jsin theta]#=#z# [ from .....#[1]#, so #dz/[d theta#=#jz#.........#[2]#
Inverting both sides of .....#[2]#, #d theta/[dz#=#1/[jz# and integrating w.r.t #z#

#theta=1/jint1/zdz# = #1/jlnz+c#, but from .....#[1]# when #theta =0,cos theta=1,sin theta=0, i.e., z=1#

Therefore, #theta=1/jln[1]+c#, but #ln [1] =0# to any base, therefore #c=0#.

So we have, #theta=1/jlnz#, i.e, #jtheta=lnx#, and so #e^j theta=z# [ theory of logs], therefore from ...#[1]#, #r[cos theta+jsin theta]=re^j theta#.......#[3]#

[ Proof of De Moivres theorem, for any value of #n#]

From .....#[3]#, #cos theta+jsin theta=e^j theta# [ omitting #r#].......#[4]#

Raising both sides of ....#[3]# to the power #n#, #[cos theta+jsin theta]^n#=#e^[n j theta]#

=#e^[j[n theta]]# , but #e^[j[n theta]]#= #cosn theta+ jsin n theta# , from .......#[4]#, replacing #theta# by # n theta#,

Therefore , #[cos theta+jsin theta]^n#= #cosn theta+jsin ntheta#.